Question

Which of the following sequence is correct for decreasing order of ionic radius?

• A. $S{e}^{-2},I^{-},B{r}^{-},O^{-2},F^{-}$
• B. $I^{-},S{e}^{-2},O^{-2},B{r}^{-},F^{-}$
• C. $S{e}^{-2},I^{-},B{r}^{-},F^{-},O^{-2}$
• D. $I^{-},S{e}^{-2},B{r}^{-},O^{-2},F^{-}$

Answer 1

The correct option is D $I^{-},S{e}^{-2},B{r}^{-},O^{-2},F^{-}$

The correct sequence for decreasing order of ionic radius is $I^{-},S{e}^{-2},B{r}^{-},O^{-2},F^{-}$.
On moving from top to bottom in a group, the ionic radii increases. Hence, $I^{-},B{r}^{-},F^{-}$.
The ionic radius of $O^{2-}$ is larger than the ionic radius of $F^{-}$ due to low effective nuclear charge which results in less attraction for the valence electrons for the nucleus.